c++cc++11undefined-behaviorunspecified-behavior

Is undefined behavior in given code?

What is the return value of f(p,p), if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value.

int f (int &x, int c) {
       c = c - 1;
       if (c==0) return 1;
       x = x + 1;
       return f(x,c) * x;
}

Options are:

  1. 3024
  2. 6561
  3. 55440
  4. 161051

I try to explain:

In this code, there will be four recursive calls with parameters (6,4), (7,3), (8,2) and (9,1). The last call returns 1. But due to pass by reference, x in all the previous functions is now 9. Hence, the value returned by f(p,p) will be 9 * 9 * 9 * 9 * 1 = 6561.

This question is from the competitive exam GATE, (see Q.no.-42). The answer key is given by GATE "Marks to all" (means there is no option correct.) key set-C, Q.no.-42. Somewhere explained as:

In GATE 2013 marks were given to all as the same code in C/C++ produces undefined behavior. This is because * is not a sequence point in C/C++. The correct code must replace

return f(x,c) * x;

with

 res = f(x,c);
 return res * x;

But the given code works fine. Is GATE's key wrong? Or is really the mistake with the question?

Solution 
  •    return f(x,c) * x;

    The result of this operation depends on the order in which the two things are evaluated. Since you cannot predict the order they'll be evaluated, you cannot predict the result of this operation.